Given the function: $\frac{s-5}{s^2-4s +5}$
The function have one zero and two poles in the RHP. So according to principle of argument (or Nyquist diagram), the Nyquist plot encircle the origin N = |Z-P| = |1-2| = 1 time. But the Nyquist plot does not show that:
What am I missing here?
It is not entirely clear what you are trying to do.
When you apply the argument principle, the contour must not pass through any of the poles or zeros.
In this case, $g(s)= {s-5 \over s^2-4s+5}$ has a zero at infinity and so the curve 'passes' through zero.
Since you are interested in the difference of right half plane poles & zeros, the contour can be modified to be $0 \to iR$, then $R e^{i\theta}$ for $\theta$ going from ${\pi \over 2}$ to $-{\pi \over 2}$ and then from $-iR$ back to zero, for some large $R$. If $R$ is sufficiently large, then the curve will pass to the right of $0$.
Note that the winding number of the right half plane poles & zeros is $-1$ with this curve, and the contour had a winding number of $+1$ around the origin, so the argument principle tells us that $1 = \sum_{z} \eta(z, \gamma) -\sum_{p} \eta(p, \gamma) = \#\text{poles} - \#\text{zeroes}$.
However...
If you are trying to apply the Nyquist stability criterion to $g$, then you are trying to determine the stability of the closed loop transfer function ${g \over 1+g}$ where $g$ is the open loop transfer function (assuming unity feedback). In this case, the system is stable if $1+g$ has no right half plane zeros.
So, either plot $1+g$ and count encirclements of zero, or plot $g$ and count encirclement of $-1$.
If this is what you are trying to go then the issue of passing through zero does not arise here.